Log NOTES Explanation Day 1

8/16/2019 Log NOTES Explanation Day 1

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Introduction To

Logarithms


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Logarithms were originallydeveloped to simplify complex

arithmetic calculations.

They were designed to transformmultiplicative processes

into additive ones.


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If at rst this seems like no big deal,then try multiplying

2,2!,!"#,#$2 and ,!"%,2!,!"#.

&ithout a calculator '

(learly, it is a lot easier to addthese two numbers.


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Indeed, they would be obsolete except for onevery important property of logarithms.

It is calledthe power property and we

will learn about it in another lesson.

*or now we need only to observe thatit is an extremely important partof solving exponential e+uations.


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ur rst -ob is to

try to makesome sense of

logarithms.


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ur rst +uestion

then must be

&hat is a logarithm /


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f course logarithms have

a precise mathematicaldenition -ust like all terms in

mathematics. )o let0s

start with that.


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1enition ofLogarithm

)uppose b3 and b4$,

there is a number 5p0such that

logb n = p if and only if b p = n


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6ow amathematician

understands exactlywhat that means.

7ut, many a

student is leftscratching theirhead.


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The rst, and perhaps

the most important step,in understandinglogarithms is to reali8e

that they always relateback to exponential

e+uations.


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9ou must be able toconvert an exponential

e+uation into logarithmicform and vice versa.

)o let0s get a lot of practice withthis '


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:xample $

)olution log2 8 = 3

&e read this as ;thelog base 2 of < is e+ual

to ;.

3Write 2 8 in logarithmic form.=


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:xample

$a

Write 42 = 16 in logarithmic form.

)olution log4 16 = 2

=ead as >the logbase ! of $% is

e+ual to 2;.


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:xample $b

)olution

Write 2− 3 =

1

8 in logarithmic form.

log21

8= − 3

1

Read as: "the log base 2 of is equal to -3".8


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kay, so now it0s timefor you to try some on

your own.

1. Write 72 = 49 in logarithmic form.

7Solution: log ! 2=


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log5 1= 0)olution

2. Write 50

= 1 in logarithmic form.


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3. Write 10− 2 =

1

100

in logarithmic form.

)olution log101

100 = − 2


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)olution log16 4 = 12

4. Finally, write 161

2= 4

in logarithmic form.


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:xample 2

Write log21

8= − 3 in exponential form.

)olution 2− 3

=

1

8


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kay, now you try thesenext three.

1. Write log10 100 = 2 in exponential form.

3. Write log27 3 =1

3 in exponential form.

2. Write log51

125

= − 3 in exponential form.


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1. Write log10 100 = 2 in exponential form.

)olution 102 = 100


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2. Write log51

125= − 3 in exponential form.

)olution 31

"

12"

−

=


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3. Write log27 3 =1

3 in exponential form.

)olution 271

3 = 3


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&e now know that alogarithm is perhaps best

understood

as beingclosely related to an

exponential e+uation.

In fact, whenever we getstuckin the problems that follow

we will return tothis one simple insight.


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&hen working with logarithms,if ever you get >stuck;, try

rewriting the problem inexponential form.

(onversely, when working

with exponential expressions,if ever you get >stuck;, try

rewriting the problemin logarithmic form.


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)olutionLet0s rewrite the

problem inexponential form.

62

= x

&e0re nished '

#Sol$e for %: log 2 x =

&%am'le 1


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)olution

5 y = 1

25

=ewrite the problemin exponential form.

Since 1

25= 5− 2

5 y

= 5− 2

y = −2

1Sol$e for (: log

2 y=

&%am'le 2


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:xample

Evaluate log3 27.

Try setting this up like this

)olution

log3 27 = y 6ow rewrite in exponentialform.3 y = 27

3 y = 33

y = 3


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These next twoproblems tend to besome of the trickiest

to evaluate.

?ctually, they are

merely identities andthe use of our simple

rulewill show this.


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:xample !

Evaluate: log7 72

)olution

6ow take it out of the logarithmic

formand write it in exponential form.

log7 72

= y*irst, we write the problem with a variable.

7 y

= 72

y = 2


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*inally, we want to take alook at the @roperty of

:+uality for Logarithmic

*unctions.Suppose b > 0 and b ≠ 1.Then logb x1 = logb x 2 if and only if x1 = x 2

7asically, with logarithmic functions,if the bases match on both sides of the

e+ual sign , then simply set thearguments e+ual.


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:xample $

Solve: log3 (4 x +10)= log3 ( x +1)

)olution)ince the bases are both 50 wesimply set the arguments e+ual.

4 x +10 = x +13 x +10 = 13 x = − 9 x = − 3


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:xample 2

Solve: log8 ( x2 −14) = log8 (5 x)

)olution)ince the bases are both 5


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:xample 2continued

Solve: log8 ( x2 −14) = log8 (5 x)

)olution x = 7 or x = −2It appears that we have 2 solutions

here.

If we take a closer look at thedenition of a logarithm however, wewill see that not only must we use

positive bases, but also we see thatthe arguments must be positive as

well. Therefore A2 is not a solution.Let0s end this lesson b takin a


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ur nal concern then is todetermine why logarithms

like the one below are

undened.

(an anyone giveus an

explanation /

2log ) 8*−


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ne easy explanation is to simplyrewrite this logarithm in exponential

form.

&e0ll then see why a negative valueis not permitted.

*irst, we write the problem with a variable.

2 y = − 8 6ow take it out of the logarithmic

formand write it in exponential form.&hat power of 2 would gives us A< /

23

= 8 and 2− 3

=

1

8

Bence expressions of this type are

undened.

2log ) 8* undefined W+,− =

2log ) 8* y− =


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That concludes our introductionto logarithms. In the lessons to

follow we will learn some important

properties of logarithms.

ne of these properties will giveus a very important tool

whichwe need to solve exponential

e+uations. Cntil then let0spractice with the basic themes

of this lesson.

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Log NOTES Explanation Day 1